Wikipedia:Reference desk/Archives/Mathematics/2011 January 16

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January 16

Bases and spans for the space of infinite sequences $\ell _{2}$

Hello everyone, I'm stuck on a homework question and could use some help.

I've been asked to solve the following:

Suppose $\{e_{i}\}_{i\in \mathbb {Z} }$ is an orthonormal basis for $\ell _{2}$, and $\forall i,\,f_{i}=e_{i}-e_{i-1}$. Show that the linear span of $\{f_{n}\}_{n\in \mathbb {Z} }$ is dense in $\ell _{2}$. Is the same true if a single element is removed from $\{f_{n}\}_{n\in \mathbb {Z} }$? What if two elements are removed?

Now I believe I've done the first part, it is the second and third which confuse me. I presume judging by the wording of the question it is still true with 1 element, otherwise the final part would be redundant. However, I can't seem to figure out why, or indeed whether the last part also fails. Could anyone help me? I've tried playing around with the algebra for a good hour or two but haven't got anywhere, so any aid would be appreciated. Many thanks in advance, Spalton232 (talk) 00:56, 16 January 2011 (UTC)

Can the sketch the solution you have found for the first part, please? The one that comes to mind for me would generalize immediately to the second situation.

Hint. In the second and third part, the entire space breaks into a direct sum of two or three subspaces, such that each of your remaining $f_{n}$ vectors lie in one of the subspaces. In the third part, one of the three subspaces has finite dimension. End hint. –Henning Makholm (talk) 03:01, 16 January 2011 (UTC)

It may be wrong but what I did was say that since $\{e_{i}\}_{i\in \mathbb {Z} }$ is a basis, suppose k is the point in $\ell _{2}$ we want to show we can be arbitrarily close to via the linear span of $\{f_{n}\}_{n\in \mathbb {Z} }$: then for some $\{a_{i}\}_{i\in \mathbb {Z} }$, we can write $k=\sum _{i\in \mathbb {Z} }a_{i}e_{i}$. Attempt to equate this to something in the linear span of $\{f_{n}\}_{n\in \mathbb {Z} }$: $\sum _{i\in \mathbb {Z} }a_{i}e_{i}=\sum _{i\in \mathbb {Z} }b_{i}f_{i}$. Then we can obtain 2 equations for the $b_{i}$: one for $b_{i+1}$ in terms of lower 'i' terms, one for $b_{i-1}$ in terms of higher 'i' terms, then calculate outwards from $b_{0}$ to find the appropriate $b_{i}$. My claim was then that $\lim _{n\to \infty }\sum _{-n}^{n}b_{i}f_{i}$ comes arbitrarily close to k. However, having written it out now, I'm concerned that under whatever norm we use, $\sum _{-n}^{n}b_{i}f_{i}$ doesn't necessarily tend to $\sum _{-\infty }^{\infty }b_{i}f_{i}$, so sadly my proof for the first part may be scuppered... Spalton232 (talk) 04:30, 16 January 2011 (UTC)

Yes, that probably won't work. One problem is that your $b_{i}$'s are not well defined. For example if k is the zero vector, then setting every $b_{i}$ to 1 solves all of your equations, but that does not have any nice convergence properties. New hints:

(a) "whatever norm we use" is implicit in the stipulation that $\{e_{i}\}_{i}$ is an orthonormal basis -- that is $\textstyle ||k||={\sqrt {\sum _{i}a_{i}^{2}}}$.

(b) Do you already know a dense subspace whose elements might be easier to approximate?

(c) You may find it simpler to prove that an arbitrary ball in $\ell ^{2}$ intersects span{f_i} than to write down an explicit expression for a sequence element that converges toward its center.

–Henning Makholm (talk) 05:14, 16 January 2011 (UTC)

My approach would be to approximate the e_i's themselves by finite linear combinations of the f_i's. If this can be done, then the problem will be solved. (And if it cannot be done, then the subspace is not dense). I think this is what Henning's hint (b) was about. To do this, it may help to note that those vectors which are finite linear combinations of the f_i's can be characterized simply in terms of their coordinates with respect to the basis (e_i). 82.124.226.199 (talk) 10:05, 16 January 2011 (UTC)

Analytic geometry question

the point (4,-2) and (-2,-5) lie on the line x÷a+y÷b=1,find the values of a and b —Preceding unsigned comment added by 41.204.142.60 (talk) 10:00, 16 January 2011 (UTC)

This appears to me to be a homework question. If you want help, you should show you've attempted to answer the question first. 82.124.226.199 (talk) 10:33, 16 January 2011 (UTC)

Hint: the line on which the points lie crosses the x axis at (a, 0), and crosses the y axis at (0, b). Gandalf61 (talk) 11:31, 16 January 2011 (UTC)

Geostationary Satellite coordinate transformation

If one placed a geostationary satellite in orbit of radius d about a body of radius r, directly above the prime meridian of the body. The satellite then fires a laser beam, the beams direction is defined by the parameters a and b. a is the vertical beam angle, the angle between the beam and equatorial plane, when the beam is projected onto the vertical plane (which is defined as containing the satellite and the axis of rotation of the body). And b is the horizontal beam angle, defined as the angle between the beam and the vertical plane when the beam is projected onto the equatorial plane.

How would one then relate the coordinates of the laser dot on the surface of the planet in latitude,longitude to the direction of the laser beam defined by a,b.

Thank you for your assistance. —Preceding unsigned comment added by 129.67.37.227 (talk) 12:57, 16 January 2011 (UTC)

The equations get pretty hairy, but I think they simplify a bit if you formulate in terms of Direction cosine. So let α, β, γ be the cosines of the angles between the beam and the x, y, and z axes reps. Putting the initial point at (-d, 0, 0), the beam is parameterized by x = -d+αt, y = βt, z = γt. To find the points of intersection with the sphere you must solve the quadratic equation t²-2dαt+d²-r²=0. Let s=√{r²-d²(1-α²)), the beam misses the sphere if the quantity under the radical is negative. Then t = dα-s is the first point of intersection, the second point being irrelevant since the beam would have to travel through the sphere to reach it. You can get x, y and z by plugging this back into the parametric equations and the direction cosines of the point with respect to the center of the sphere would be x/r, y/r, z/r.

This is assuming that the celestial body is a sphere and perfectly smooth, which is only true as an approximation.--RDBury (talk) 15:55, 16 January 2011 (UTC)

"General Limit Theory"

Hello. How often is the following theory regarding to finding limits of some argument A by direct substitution valid?

1. If it equals some real number, then that is the limit L.
2. If it equals some non-zero number over zero, then the limit does not exist.
3. In most other cases, apply l'Hôpital's rule.

Applying l'Hôpital's rule to $\lim _{x\to \infty }{\frac {x}{\sqrt {x^{2}+1}}}$ yields the same argument. That can be easily solved by factoring x from the denominator. Are there any other exceptions? Thanks in advance. --Mayfare (talk) 17:04, 16 January 2011 (UTC)

I think you mean "quotient" when you write "argument". Have you looked at the caveats and exceptions in our L'Hôpital's rule article? –Henning Makholm (talk) 17:35, 16 January 2011 (UTC)

magic machine statistics

In a casino you find a magic machine that will take tokens in any denomination from $1, $10, $100, $1000, $10k, $100k, $1m, $10m, $100m, $1b, $10b, $100b. However, you may only play with one kind of denomination.

The machine works like this: you put a token in one at a time, and with 7/8th probability it eats it - you get nothing. However, the remaining 1/8th probability brings the expected value above 1: namely, the payoff is 10:1 (instead of 8:1 -- or is it 7:1 ? -- as it would have to be to make the bet an even game). You can play as long as you like, but obviously limited to the order of hours, so if you start with $1 you will never make it to $1b (you would physically have to put in a hundred million plus chips, and you don't have time or stamina to do that). Then they will take the machine away forever. You can't upgrade to bigger tokens once you've started.

Now you're at home, writing the Wikipedia Reference Desk, to see how much of a bankroll you should put together given your means and connections and risk-aversion. (You will sell your house to use this machine, so obviously you're not going to be happy if you come with a single $1m chip, put it in, and lose it -- the probability of which is 7/8, if that's your single chip. The fact that there was a positive expected value doesn't help you in this case...)

It seems to me that the Reference desk's answer should hold just as easily for $1, $1k, $1m, and $1b chips. So let me just ask in terms of $1 (you can correct my misunderstanding if I am wrong that it applies equally to any other denomination).

Let me tell you what I know. If you walk in with a single $1 chip and put it in, you have an 87.5+% chance of losing your opportunity to take advantage of the positive expected return on each play!!! (This is the 7/8th probability of its eating your coin on the first play. The remaining "+" that I put includes the same calculation for the second and subsequent rounds...).

Obviously this is unacceptable. This is a machine that gives you a positive expected value on every single play!!! I want eight sigmas of guarantee that I will not lose my chance to exploit it.

So, the question is simply this: for every sigma between 1 and 7, how many times the $1 chip should I have with me to ensure that the chances of losing it all are that unlikely?

Here is a table of values I'm interested in. sigmas / = percentage confidence I'll have in not losing all my chips before being "in the clear":

0.674σ 	50%
1σ 	68.2689492%
1.645σ 	90%
1.960σ 	95%
2σ 	95.4499736%
2.576σ 	99%
3σ 	99.7300204%
3.2906σ 99.9%
4σ 	99.993666%
5σ 	99.9999426697%
6σ 	99.9999998027%
7σ 	99.9999999997440%

So the kind of answer I'm looking for is, 'yeah, come in with 4 chips for 1 sigma, 7 for 2, 15 for 3, 18 for 4, 22 for 5, 27 for 6, 30 for 7 sigmas. if you come in with 30 chips the chances that you will NOT lose them all in the course of the next 5000 rolls is 99.9999999997440%.' (these numbers are just made up -- it's what I imagine the format of the answer would be like).

however if there are any uncertainties in my question I'm happy to add more information -- and if I have drastically misunderstood something in statistics, I'm sorry...!!! Please explain gently why my question is meaningless, if it is.... 194.78.208.19 (talk) 18:48, 16 January 2011 (UTC)

First note that this translation table between standard scores and probabilities is only applicable for the normal distribution. Here you don't have a normal distribution and sigmas, just plain old probabilities.

Now let's denote by $p=1/8$ the probability of winning a round, $K=10$ the number of chips you end up with after successfully betting one, and $f(n)$ the probability of going broke starting with n chips. The function satisfies $f(n+1)=f(n)f(1)$ (since losing $n+1$ chips means losing the first n and then losing the remaining one) and thus $f(n)=f(1)^{n}$. It also satisfies $f(1)=1-p+pf(K)$, since losing a chip means either losing it right away or winning and then losing the K chips you get. So the equation you need to solve is $pf(1)^{K}-f(1)=p-1$. In general your best bet for this is to solve it numerically, and for your parameters you get $f(1)=0.9492654...$. So to get >0.999999999997440 probability of winning you need $n=513$. For a probability of q you need $n={\frac {\log(1-q)}{\log(f(1))}}$.

Note that in real life, I'd approach this by considering the utility of money, which can be reasonably modeled as a logarithmic function. A bet that has a positive expected payoff can have a negative expected utility if the variance is too high. By having more chips you reduce the variance until the expected utility is positive. -- Meni Rosenfeld (talk) 19:28, 16 January 2011 (UTC)

Thank you -- I know I'm "reducing the variance until the expected utility is positive" :). I'm afraid I have very little training in mathematics, so I'm having trouble following your formulas, however I notice you gave me a fine and concrete answer for what above I erroneously called 7 sigmas: your answer is "513 chips." Could you give me the same number as corresponding to the other erronesouly called "sigmas". i.e. could yuo give me the number of chips that would result in, respectively,

a 68.2689492%
b 90%
c 95.4499736%
d 99.73%
e 99.993666%
f 99.9999426697% and
g 99.9999998027

percent probability that I will not lose all the chips within the first few thousand plays? Thanks for your fast answer!! 194.78.208.19 (talk) 19:54, 16 January 2011 (UTC)

Note: the reason it is THESE exact probabilities that interest me is that I have a LOT of experience with data following the normal curve! Everything from IQ's to Heights, I can easily imagien in terms of the mean for the population and the known standard deviation, I have a huge body of intutiive reference for many many things that follow that curve and for which I know the mean and standard deviation. So, even though this particular data might not follow the curve, it is very helpful for me to be able to put the SAME percentages in those terms... 194.78.208.19 (talk) 20:00, 16 January 2011 (UTC)

That would be

a 23
b 45
c 60
d 114
e 186
f 277
g 385

Note that my calculations assume the game goes on forever (as long as you have chips) rather than stopping after some iterations, but that shouldn't have a significant effect on the results. -- Meni Rosenfeld (talk) 20:07, 16 January 2011 (UTC)

Thanks so much! When you say "not a signfiicant effect", could you specify what you mean? For example, do you have a way of verifying numerically how the probabilities change if you stop playing when you've made 10x your money? (if you walk in with 23 chips, you will leave if you hit 230). These are huge numbers, by the way. For example, with e, what I would like to play with if my house is on the line, 186 chips means that if my house is worth $1m and I sell it to play, I can only play with $1k chips! (I can't raise enough money to afford 186 $10k chips). If it is not an inordinate amount of work for you, I wonder if you could also reply with the numbers for the 1/8th payoff increasing from 10:1 to 12:1 or to 16:1? Are either of those enough to push the number of required chips down to something reasonable? Really anything less than d, with 114 chips, is just plain irresponsible given the machine. Why would you chance even a 5% chance of losing all your money (i.e. c above, at 60 chips) when you are guaranteed limitless positive bets??? What does the number change to from 114 if the winning bet goes up to 12:1 or 16:1 payoff? THanks! 194.78.208.19 (talk) 20:16, 16 January 2011 (UTC)

Let me write what I just wrote again: if you have a chance to play this game (you are guaranteed that the conditions are as written above) it is just plain irresponsible of you to play with anything less than 99.73% probability that you won't just lose your money outright. Actually that's pushing it: you should try to have 99.993% chance of it. Playing with a 95% chance of not losing yoru money is so irresponsible. You are taking a one in twenty chance of losing everything you've worked for. I guess now I can understand why companies like Apple keep 20 billion in cash around: they have the chance to play in something with positive expectation, but due to the vagaries of the world it is just plain irresponsible of them to risk everything they've worked for on a few bad stochastic seasons/reasons... 194.78.208.19 (talk) 20:21, 16 January 2011 (UTC)

A better option with a $1M house might be to take $10k chips but leave some cash on the side. Anyway, you should be able to plug in the formulas any values you'd like, Wolfram Alpha can help you solve the required equation for different p and K -- Meni Rosenfeld (talk) 09:28, 17 January 2011 (UTC)

THanks, I'll try to get Wolfram Alpha to do that. (All the same, if you do have time at some point, I would appreciate the same data you just gave me, but for 12:1 and 16:1 payoffs on the 1/8th probability (with loss of chip in the other 7/8th). However, I am puzzled by your statement "A better option with a $1M house might be to take $10k chips but leave some cash on the side." When you have a house you've put a million dollars into, you will not be able to look yourself in the mirror if you just lost it, because you went with d (114 chips) at 99.73% chance of winning, or fewer chips. 99.73% chance is just not enough. To give an example, that's practically how much my own SAT score was above the mean for all Amwericans. I know it's not THAT rare, and I could never live with myself if I lost my house because I was ok with just having a 99.73% chance of winning. THat's why I pick the next level, e: 99.993666% chance of winning. I don't know anyone who's at 99.993666% level at ANYTHING in the general population (I don't know anyone who's 8 feet, or has an IQ of 200, or even got a perfect score on their SAT's, which is not that rare). Now for e, you told me I need 186 chips. But the problem is this:

1 million / 186 = 5 376. So I could buy 5k chips - but there aren't any! There are 10, 100, 1k, 10k, and 100k chips. At the next level, 10k chips, I could only buy 100 chips - a totally unacceptable risk of losing what I've worked for my whole life. I might as well commit suicide. Would you take between a 1 in 20 and 1 in 100 chance of shooting yourself in the head? Of course not: it's irresponsible. Would you take a 99.993666% of not shooting yourself in the head, if you would end up fabulously wealthy? I would take that risk. I will take my 0.0063% chance of losing my house, and if I do, then I will chalk it up to bad luck, and not irresponsibility. That's why I HAVE to have 186 chips, the 100 chips I can buy at 10k is just not enough. So, the way I'm reading your last comment, it only makes sense if you mistyped "10k" when you meant "1k"... Anyway thanks for all your help so far! 194.78.208.19 (talk) 12:00, 17 January 2011 (UTC)

As I said, I would approach this by considering the utility of money rather than my emotional attachment to my house. Going with $10K chips will most probably leave me with 10 times the money as $1K chips, and I'll take it unless the risk is too high.

Whether the risk is high enough depends on the particular numbers. I submit that for the given numbers, going with $1K chips may be the better option after all. -- Meni Rosenfeld (talk) 12:26, 17 January 2011 (UTC)

So you find someone else who also has a $1m house, combine your funds to buy 200 $10k chips, then play until you are exhausted and split your winnings 50:50. Problem then is how can you be certain that whoever runs the machine will pay up - if someone offered this good a deal in real life, it would certainly be a scam. Gandalf61 (talk) 12:44, 17 January 2011 (UTC)

Well, why find someone with a house worth 1m, when the Chinese government has 1.2 trillion us dolalrs just lying around? (The definition of "foreign currency reserves"). If I divide that by 385 chips, it is $3.1 billion, so I could just use 385 $1b chips by using the Chinese government's moeny -- also, they would still have more than 2/3 of their money in case they need it in liquid form for other reasons (are they blackmailing the US with it by threatening to just dump it on the market? are they manipulating their exchange rates with it in an active way? etc.).

However, I don't expect to be able to convince either the Chinese government, or even Warren Buffett, or even my neighbor. In fact, I don't believe in doing so. Whatever my personal resources are, that's what I plan on using, that is what I consider "right" to use, leverage be damned. As for your other concern, whether the "magic machine" will play up, that's the beauty of two thigns. One, the word "magic", and two, the word "Mathematics" at the top of this page. It is not a Science or Humanities question - I am simply interested in the math of it! :) By the way,

Gandalf, if you understand Meni's derivation/formulas above, maybe you could use them with WolframAlpha to come up with the same a-f figures:

“

a 68.2689492%
b 90%
c 95.4499736%
d 99.73%
e 99.993666%
f 99.9999426697% and
g 99.9999998027

”

that he did, but where the payoff is 12:1 or 16:1 respectively instead of 10:1 when you get the winning play? If you will just calculate "f(1)" for me for these two other possible values, I can plug the rest into wolfram alpha... I'm having trouble doing it myself. thanks! 194.78.208.19 (talk) 13:01, 17 January 2011 (UTC)

For p = 1/8 and K = 12 I get f(1) to be approximately 0.9224465, and for K = 16 I get approximately 0.8969307. Gandalf61 (talk) 13:25, 17 January 2011 (UTC)